College Trigonometry, 2011a

928 Applications of Trigonometry
information by dividing both sides of r cos 2 (θ) =sin(θ) bycos 2 (θ). Doing so, we get
r = sin(θ) ,orr =sec(θ) tan(θ). As before, the r = 0 case is recovered in the solution
cos 2 (θ)
r = sec(θ) tan(θ) (letθ = 0), so we state the latter as our final answer.
2. As a general rule, converting equations from polar to rectangular coordinates isn’t as straight
forward as the reverse process. We could solve r 2 = x 2 + y 2 for r to get r = ± √ x 2 + y 2
and solving tan(θ) = y x
requires the arctangent function to get θ = arctan( y
x)
+ πk for
integers k. Neither of these expressions for r and θ are especially user-friendly, so we opt for
a second strategy – rearrange the given polar equation so that the expressions r 2 = x 2 + y 2 ,
r cos(θ) =x, r sin(θ) =y and/or tan(θ) = y x
present themselves.
(a) Starting with r = −3, we can square both sides to get r 2 =(−3) 2 or r 2 =9. Wemaynow
substitute r 2 = x 2 + y 2 to get the equation x 2 + y 2 = 9. As we have seen, 9 squaring an
equation does not, in general, produce an equivalent equation. The concern here is that
the equation r 2 = 9 might be satisfied by more points than r = −3. On the surface, this
appears to be the case since r 2 = 9 is equivalent to r = ±3, not just r = −3. However,
any point with polar coordinates (3,θ) can be represented as (−3,θ+ π), which means
any point (r, θ) whose polar coordinates satisfy the relation r = ±3 has an equivalent 10
representation which satisfies r = −3.
to get tan(θ) =tan( ) √
4π
3 = 3.
(b) We take the tangent of both sides the equation θ = 4π 3
Since tan(θ) = y x ,weget y x = √ 3ory = x √ 3. Of course, we pause a moment to wonder
if, geometrically, the equations θ = 4π 3 and y = x√ 3 generate the same set of points. 11
The same argument presented in number 1b applies equally well here so we are done.
(c) Once again, we need to manipulate r =1− cos(θ) a bit before using the conversion
formulas given in Theorem 11.7. We could square both sides of this equation like we did
in part 2a above to obtain an r 2 on the left hand side, but that does nothing helpful for
the right hand side. Instead, we multiply both sides by r to obtain r 2 = r − r cos(θ).
We now have an r 2 and an r cos(θ) in the equation, which we can easily handle, but
we also have another r to deal with. Rewriting the equation as r = r 2 + r cos(θ)
and squaring both sides yields r 2 = ( r 2 + r cos(θ) ) 2 . Substituting r 2 = x 2 + y 2 and
r cos(θ) = x gives x 2 + y 2 = ( x 2 + y 2 + x ) 2 . Once again, we have performed some
9 Exercise 5.3.1 in Section 5.3, for instance . . .
10 Here, ‘equivalent’ means they represent the same point in the plane. As ordered pairs, (3, 0) and (−3,π)are
different, but when interpreted as polar coordinates, they correspond to the same point in the plane. Mathematically
speaking, relations are sets of ordered pairs, so the equations r 2 =9andr = −3 represent different relations since
they correspond to different sets of ordered pairs. Since polar coordinates were defined geometrically to describe the
location of points in the plane, however, we concern ourselves only with ensuring that the sets of points in the plane
generated by two equations are the same. This was not an issue, by the way, when we first defined relations as sets
of points in the plane in Section 1.2. Back then, a point in the plane was identified with a unique ordered pair given
by its Cartesian coordinates.
11 In addition to taking the tangent of both sides of an equation (There are infinitely many solutions to tan(θ) = √ 3,
and θ = 4π is only one of them!), we also went from y = √ 3, in which x cannot be 0, to y = x √ 3 in which we assume
3
x
x can be 0.